Unformatted text preview: d i . #2. Make a guess that a solution Y i = Y i ( t ) of the nonhomogeneous equation n ¹ m =1 P n − m ( t ) Y ( m ) i ( t ) = G i ( t ) for i = 1 , 2 , . . . , r ; is in the form Y i ( t ) = d i + n ¹ j =0 A ij t j e α i t cos β i t + d i + n ¹ j =0 B ij t j e α i t sin β i t for some constants A ij and B ij . Note that α i , and β i are the same as above, and that Y i ( t ) involves a polynomial of degree ( d i + n ). #3. Recombine as the sum Y ( t ) = Y 1 ( t ) + Y 2 ( t ) + · · · + Y r ( t ); then Y = Y ( t ) is the desired solution to the nonhomogeneous equation a d n Y dt n + a 1 d n − 1 Y dt n − 1 + · · · + a n − 1 dY dt + a n Y = G ( t ) ....
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.
 Spring '09
 EdrayGoins

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